a) 2Al +3CuSO4 --> Al2(SO4)3 + 3Cu (1)
nCuSO4=0,125(mol)
theo (1) nAl=2/3nCu=1/12(mol)
=>mAl=2,25(g)
b) theo (1) : nAl2(SO4)3=1/3nCuSO4=1/24(mol)
=>mAl2(SO4)3=14,25(g)
nCu=nCuSO4=0,125(mol)
=>mCu=8(g)
mdd sau pư=2,25+100-8=94,25(g)
=>C%dd Al2(SO4)3=15,12(%)